• question_answer 10) Permanganate (VII) ion, $MnO_{4}^{-}$ in basic solution oxidises iodide ion, ${{I}^{-}}$ to produce molecular iodine $({{I}_{2}})$ and manganese (IV) oxide$(Mn{{O}_{2}})$. Write the balanced ionic equation to represent this redox reaction.

Reaction: $MnO_{4}^{-}(aq)+{{I}^{-}}(aq)\to Mn{{O}_{2}}(s)+{{I}_{2}}(s)$ 1st step: Splitting into two half reactions: (balance the elements other than hydrogen and oxygen) $MnO_{4}^{-}\to Mn{{O}_{2}}(s)$ (Reduction) $2{{I}^{-}}\to {{I}_{2}}(s)$ (Oxidation) 2nd step: Balance oxygen atoms by adding water molecule to the oxygen deficient side. $MnO_{4}^{-}\to Mn{{O}_{2}}+2{{H}_{2}}O$ $2{{I}^{-}}\to {{I}_{2}}$ 3rd step: Balance hydrogen by adding ${{H}^{+}}$ ions to the hydrogen deficient side. $MnO_{4}^{-}+4{{H}^{+}}\to Mn{{O}_{2}}+2{{H}_{2}}O$ $2{{I}^{-}}\to {{I}_{2}}$ 4th step: Balance the charge by adding electrons $MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O$ $2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}$ 5th step: Equations are added in such a way that electrons are cancelled; we now get the balanced equation. $[MnO_{4}^{-}]+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O]\times 2$ $[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}]\times 3$ $\frac{\frac{{}}{2MnO_{4}^{-}+6{{I}^{-}}+8{{H}^{+}}\to 2Mn{{O}_{2}}+3{{I}_{2}}+4{{H}_{2}}O}}{Balanced\text{ }equation}$ $2MnO_{4}^{-}+4{{H}_{2}}O\to 2Mn{{O}_{2}}+3\underset{(Basic\,medium)}{\mathop{{{I}_{2}}+8O{{H}^{-}}}}\,$